Question: You have found the following ages (in years) of all 6 bears at your local zoo: $ 42,\enspace 12,\enspace 24,\enspace 5,\enspace 22,\enspace 21$ What is the average age of the bears at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 6 bears at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{42 + 12 + 24 + 5 + 22 + 21}{{6}} = {21\text{ years old}} $ Find the squared deviations from the mean for each bear. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $42$ years $21$ years $441$ years $^2$ $12$ years $-9$ years $81$ years $^2$ $24$ years $3$ years $9$ years $^2$ $5$ years $-16$ years $256$ years $^2$ $22$ years $1$ year $1$ year $^2$ $21$ years $0$ years $0$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{441} + {81} + {9} + {256} + {1} + {0}} {{6}} $ $ {\sigma^2} = \dfrac{{788}}{{6}} = {131.33\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{131.33\text{ years}^2}} = {11.5\text{ years}} $ The average bear at the zoo is 21 years old. There is a standard deviation of 11.5 years.